Find the value of $a$ so that the lines described by
\[\begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + t \begin{pmatrix} a \\ -2 \\ 1 \end{pmatrix}\]and
\[\begin{pmatrix} 1 \\ -3/2 \\ -5 \end{pmatrix} + u \begin{pmatrix} 1 \\ 3/2 \\ 2 \end{pmatrix}\]are perpendicular.
Explanation: The direction vector of the first line is $\begin{pmatrix} a \\ -2 \\ 1 \end{pmatrix}.$  The direction vector of the second line is $\begin{pmatrix} 1 \\ 3/2 \\ 2 \end{pmatrix}.$

The lines are orthogonal when the direction vectors will be orthogonal, which means their dot product will be 0.  This gives us
\[(a)(1) + (-2) \left( \frac{3}{2} \right) + (1)(2) = 0.\]Solving, we find $a = \boxed{1}.$